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Metamath Proof Explorer

Theorem lsmss1

Description: Subgroup sum with a subset. (Contributed by NM, 27-Mar-2014) (Revised by Mario Carneiro, 19-Apr-2016)

Ref Expression
Hypothesis lsmub1.p ⊢ ⊕ = ( LSSum ‘ 𝐺 )
Assertion lsmss1 ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑇 ⊆ 𝑈 ) → ( 𝑇 ⊕ 𝑈 ) = 𝑈 )

Proof

Step Hyp Ref Expression
1 lsmub1.p ⊢ ⊕ = ( LSSum ‘ 𝐺 )
2 ssid ⊢ 𝑈 ⊆ 𝑈
3 1 lsmlub ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ) → ( ( 𝑇 ⊆ 𝑈 ∧ 𝑈 ⊆ 𝑈 ) ↔ ( 𝑇 ⊕ 𝑈 ) ⊆ 𝑈 ) )
4 3 3anidm23 ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ) → ( ( 𝑇 ⊆ 𝑈 ∧ 𝑈 ⊆ 𝑈 ) ↔ ( 𝑇 ⊕ 𝑈 ) ⊆ 𝑈 ) )
5 4 biimpd ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ) → ( ( 𝑇 ⊆ 𝑈 ∧ 𝑈 ⊆ 𝑈 ) → ( 𝑇 ⊕ 𝑈 ) ⊆ 𝑈 ) )
6 2 5 mpan2i ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ) → ( 𝑇 ⊆ 𝑈 → ( 𝑇 ⊕ 𝑈 ) ⊆ 𝑈 ) )
7 6 3impia ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑇 ⊆ 𝑈 ) → ( 𝑇 ⊕ 𝑈 ) ⊆ 𝑈 )
8 1 lsmub2 ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ) → 𝑈 ⊆ ( 𝑇 ⊕ 𝑈 ) )
9 8 3adant3 ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑇 ⊆ 𝑈 ) → 𝑈 ⊆ ( 𝑇 ⊕ 𝑈 ) )
10 7 9 eqssd ⊢ ( ( 𝑇 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑈 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑇 ⊆ 𝑈 ) → ( 𝑇 ⊕ 𝑈 ) = 𝑈 )